列表中按组件代码

时间:2016-06-15 11:16:03

标签: python list group-by max

我想从字典列表中提取按其代码分组的最大运行日期的所有元素。 这是我到目前为止所得到的:

import datetime
from itertools import groupby

commission_list = [
    {'code': 'COMMISSION_CODE1', 'runningdt': datetime.datetime.strptime('2016-04-12', '%Y-%m-%d'), 'value': 150},
    {'code': 'COMMISSION_CODE1', 'runningdt': datetime.datetime.strptime('2016-04-12', '%Y-%m-%d'), 'value': 450},
    {'code': 'COMMISSION_CODE1', 'runningdt': datetime.datetime.strptime('2016-04-16', '%Y-%m-%d'), 'value': 140},
    {'code': 'COMMISSION_CODE1', 'runningdt': datetime.datetime.strptime('2016-04-17', '%Y-%m-%d'), 'value': 120},
    {'code': 'COMMISSION_CODE1', 'runningdt': datetime.datetime.strptime('2016-04-17', '%Y-%m-%d'), 'value': 220},

    {'code': 'COMMISSION_CODE2', 'runningdt': datetime.datetime.strptime('2016-04-11', '%Y-%m-%d'), 'value': 150},
    {'code': 'COMMISSION_CODE2', 'runningdt': datetime.datetime.strptime('2016-04-15', '%Y-%m-%d'), 'value': 140},
    {'code': 'COMMISSION_CODE2', 'runningdt': datetime.datetime.strptime('2016-04-16', '%Y-%m-%d'), 'value': 160},
    {'code': 'COMMISSION_CODE2', 'runningdt': datetime.datetime.strptime('2016-04-19', '%Y-%m-%d'), 'value': 210},

    {'code': 'COMMISSION_CODE3', 'runningdt': datetime.datetime.strptime('2016-04-16', '%Y-%m-%d'), 'value': 330},
    {'code': 'COMMISSION_CODE3', 'runningdt': datetime.datetime.strptime('2016-04-20', '%Y-%m-%d'), 'value': 310},
    {'code': 'COMMISSION_CODE3', 'runningdt': datetime.datetime.strptime('2016-04-20', '%Y-%m-%d'), 'value': 410},
]

latest_run_commissions = []
for key, commission_group in groupby(commission_list, lambda x: x['code']):
    tem = list(commission_group)
    the_last_com = (max(tem, key=lambda x: x['runningdt']))
    filtered_objs = filter(lambda f: f['runningdt'] == the_last_com['runningdt'], tem)
    for o in filtered_objs:
        latest_run_commissions.append(o)

for f in latest_run_commissions:
    print(f)
print(" ")

还有更有效率的方法吗?您的建议或建议将受到高度赞赏和欢迎。

1 个答案:

答案 0 :(得分:0)

您可以使用SELECT *,FIND_IN_SET( mark , ( SELECT GROUP_CONCAT( mark ORDER BY mark DESC ) FROM exam_score ) ) AS rank FROM exam_score ORDER BY rank asc 模块中的itemgetter来有效地执行此操作。

operator